Question

PLEASE SOLVE THIS FRIENDS I BEG U​

Answer 1

Given:

• A line segment joining the point (-3,10) and (6,-8)
• A point (-1,6) divides the line segment in a certain ratio

To find:

• Ratio in which the given point will divide the line segment

Formula required:

• Section formula

$\purple{\bigstar}\;\;\;\boxed{\sf{(x,\;y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}}$

[ Where point $\sf{(x,\;y)}$ is dividing the line segment joining the points $\sf{(x_1,\;y_1)}$ and $\sf{(x_2,\;y_2)}$ in the ratio m : n ]

Solution:

On comparison we will get,

• $\sf{x_1=-3,\;y_1=10}$

• $\sf{x_2=6,\;y_2=-8}$

• $\sf{x=-1,\;y=6}$

Now,

Let the given point divide the line segment in the ratio m : n

so, Let us assume that

$\sf{\dfrac{m}{n}=\dfrac{k}{1}}$

then,

Using section formula

$\implies\sf{(x,\;y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}$

$\implies\sf{(-1,\;6)=\bigg(\dfrac{(k)(6)+(1)(-3)}{k+1}\;,\;\dfrac{(k)(-8)+(1)(10)}{k+1}\bigg)}$

$\implies\sf{-1=\dfrac{(k)(6)+(1)(-3)}{k+1}\;,\;6=\dfrac{(k)(-8)+(1)(10)}{k+1}}$

$\implies\sf{-k-1=6k-3\;\;,\;\;6k+6=-8k+10}$

$\implies\sf{7k=2\;\;,\;\;14k=4}$

From here we will get,

$\implies\boxed{\sf{k=\dfrac{2}{7}=\dfrac{m}{n}}}$

Hence,

• m : n = 2 : 7

And therefore,

• Given point will divide the given line segment in the ratio 2 : 7.

Answer 2

Given:

A line segment joining the point (-3,10) and (6,-8)

A point (-1,6) divides the line segment in a certain ratio

To find:

Ratio in which the given point will divide the line segment

Formula required:

Section formula

\purple{\bigstar}\;\;\;\boxed{\sf{(x,\;y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}}★

(x,y)=(

m+n

mx

2

+nx

1

,

m+n

my

2

+ny

1

)

[ Where point \sf{(x,\;y)}(x,y) is dividing the line segment joining the points \sf{(x_1,\;y_1)}(x

1

,y

1

) and \sf{(x_2,\;y_2)}(x

2

,y

2

) in the ratio m : n ]

Solution:

On comparison we will get,

\sf{x_1=-3,\;y_1=10}x

1

=−3,y

1

=10

\sf{x_2=6,\;y_2=-8}x

2

=6,y

2

=−8

\sf{x=-1,\;y=6}x=−1,y=6

Now,

Let the given point divide the line segment in the ratio m : n

so, Let us assume that

\sf{\dfrac{m}{n}=\dfrac{k}{1}}

n

m

=

1

k

then,

Using section formula

\implies\sf{(x,\;y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}⟹(x,y)=(

m+n

mx

2

+nx

1

,

m+n

my

2

+ny

1

)

\implies\sf{(-1,\;6)=\bigg(\dfrac{(k)(6)+(1)(-3)}{k+1}\;,\;\dfrac{(k)(-8)+(1)(10)}{k+1}\bigg)}⟹(−1,6)=(

k+1

(k)(6)+(1)(−3)

,

k+1

(k)(−8)+(1)(10)

)

\implies\sf{-1=\dfrac{(k)(6)+(1)(-3)}{k+1}\;,\;6=\dfrac{(k)(-8)+(1)(10)}{k+1}}⟹−1=

k+1

(k)(6)+(1)(−3)

,6=

k+1

(k)(−8)+(1)(10)

\implies\sf{-k-1=6k-3\;\;,\;\;6k+6=-8k+10}⟹−k−1=6k−3,6k+6=−8k+10

\implies\sf{7k=2\;\;,\;\;14k=4}⟹7k=2,14k=4

From here we will get,

\implies\boxed{\sf{k=\dfrac{2}{7}=\dfrac{m}{n}}}⟹

k=

7

2

=

n

m

Hence,

m : n = 2 : 7

And therefore,

Given point will divide the given line segment in the ratio 2 : 7.

Hope it helps you

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